What is the value of $\dfrac{d}{dx}\left(\sqrt{x}\right)$ at $x=9$ ?
Answer: The strategy We can first rewrite the radical as a rational power of $x$. Then, the derivative can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Once we have the derivative, we can evaluate it at $x=9$. Rewriting the radical as a rational power $\sqrt{x}=x^{^{\frac{1}{2}}}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(x^{^{\frac{1}{2}}}\right) \\\\ &=\dfrac{1}{2}x^{^{\frac{1}{2}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac12x^{^{-\frac{1}{2}}} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac12x^{^{-\frac{1}{2}}}$, which can also be written as $\dfrac{1}{2\sqrt{x}}$. Now let's plug ${x=9}$ : $\begin{aligned} \dfrac{1}{2\sqrt{{9}}}&=\dfrac{1}{2\cdot 3} \\\\ &=\dfrac16 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\sqrt{x}\right)$ at $x=9$ is $\dfrac16$.